Find $\lim_{x\to 3}\dfrac{(x-5)^2}{(x-3)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2}{3}$ (Choice B) B $0$ (Choice C) C $\dfrac{11}{3}$ (Choice D) D The limit doesn't exist
Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to 3}\dfrac{(x-5)^2}{(x-3)^2}&=\dfrac{(3-5)^2}{(3-3)^2} \\\\ &=\dfrac{(-2)^2}{0^2} \\\\ &=\dfrac{4}{0} \end{aligned}$ Our expression evaluates to a nonzero number over zero. In such cases, the limit doesn't exist. In conclusion, $\lim_{x\to 3}\dfrac{(x-5)^2}{(x-3)^2}$ doesn't exist.